What is the extraneous solution to these equations? $\dfrac{x^2 + 14x}{x - 6} = \dfrac{29x - 54}{x - 6}$
Answer: Multiply both sides by $x - 6$ $ \dfrac{x^2 + 14x}{x - 6} (x - 6) = \dfrac{29x - 54}{x - 6} (x - 6)$ $ x^2 + 14x = 29x - 54$ Subtract $29x - 54$ from both sides: $ x^2 + 14x - (29x - 54) = 29x - 54 - (29x - 54)$ $ x^2 + 14x - 29x + 54 = 0$ $ x^2 - 15x + 54 = 0$ Factor the expression: $ (x - 6)(x - 9) = 0$ Therefore $x = 6$ or $x = 9$ At $x = 6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 6$, it is an extraneous solution.